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How do I find the distribution of $Y = aX + b$ given that $X \sim N(\mu = \alpha, \sigma^{2} = \beta)$.

Let's say for example I have $X \sim N(\mu = 3, \sigma^{2} = 16)$ and want to find out the distribution of $Y = 2X - 5$ and compute some probability $P(2X-5 > 0)$, how would I go about doing this?

Does multiplying a r.v $X$ with a constant $a$ and then adding a constant $b$ change the distribution or will the new r.v $Y = 2X-5$ still be normally distributed with the same mean $\alpha$ and variance $\beta$ as $X$?

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2 Answers 2

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$Y$ still has the normal distribution since it is a linear transformation of a normally distributed random variable $X$. However, the mean and the variance are not the same. We need to calculate the expected value and the variance to determine the parameters of the distribution of $Y$. We have that $$ \operatorname E[Y]=\operatorname E[aX+b]=a\operatorname E[X]+b=a\alpha+b $$ and $$ \operatorname{Var}[Y]=\operatorname{Var}[aX+b]=a^2\operatorname{Var}[X]=a^2\beta. $$ Hence, $Y\sim N(a\alpha+b,a^2\beta)$.

This holds for normally distributed random variables but does not necessarily hold for other distributions (for example, it does not hold for the Bernoulli distribution).

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    $\begingroup$ A minor point of clarification. The fact that $Y$ remains normal under such a transformation is a property of $X$ having a normal distribution, but the mean and variance being $a \alpha + b$ and $a^2 \beta$, respectively, is true even when $X$ is not normally distributed. I just did not want the OP to think that those formulas only apply when $X$ is normal. $\endgroup$
    – heropup
    Commented Oct 19, 2021 at 18:42
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Alternatively, you can use the X variable moment-generating function since $X \sim N(\alpha, \beta)$. You only have to check $Y$ has a moment-generating function similar to a Normal distribution's one.

By definition $$m_Y(t) = E[e^{tY}]$$ $$ = E[e^{t(aX+b)}]$$ $$ = E[e^{taX}e^{tb}]$$ $$ = e^{tb}E[e^{taX}]$$ $$ = e^{tb}(m_X(at))$$ $$ = e^{tb}(e^{\alpha(at) + \frac{\beta (at)^2}{2}})$$ $$ = e^{(\alpha a) t + tb + \frac{\beta (at)^2}{2}}$$ $$ = e^{(\alpha a + b)t + \frac{(\beta a^2) t^2}{2}}$$

Let $\mu =(\alpha a + b)$ and $\sigma = (\beta a^2)$. Then $$m_Y(t) = e^{\mu t + \frac{\sigma t^2}{2}}$$

As you may see this is a Normal distribution moment-generating function. By inspection, you can check its parameters are as follows: $E[Y] = (\alpha a + b)$ and $Var[Y] = \beta a^2$.

Then $Y \sim N(\alpha a + b, \beta a^2)$.

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